How do you simplify #root4(16) *root4(128)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Lovecraft Oct 2, 2015 #root(4)(16)*root(4)(128) = 4root(4)(2^(3))# Explanation: #2*2*2*2 =16 = 2^4# #16*2*2*2 = 128 = 2^(4+3) = 2^7# #root(4)(16)*root(4)(128) = root(4)(2^4)*root(4)(2^(4+3))# #root(4)(16)*root(4)(128) = 2*2*root(4)(2^(3))# #root(4)(16)*root(4)(128) = 4root(4)(2^(3))# Answer link Related questions How do you simplify #\frac{2}{\sqrt{3}}#? How do you multiply and divide radicals? How do you rationalize the denominator? What is Multiplication and Division of Radicals? How do you simplify #7/(""^3sqrt(5)#? How do you multiply #(sqrt(a) +sqrt(b))(sqrt(a)-sqrt(b))#? How do you rationalize the denominator for #\frac{2x}{\sqrt{5}x}#? Do you always have to rationalize the denominator? How do you simplify #sqrt(5)sqrt(15)#? How do you simplify #(7sqrt(13) + 2sqrt(6))(2sqrt(3)+3sqrt(6))#? See all questions in Multiplication and Division of Radicals Impact of this question 1222 views around the world You can reuse this answer Creative Commons License